Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(=>2x+\frac{3}{5}=\frac{3}{5}\)
\(2x=\frac{3}{5}-\frac{3}{5}\)
\(2x=0\)
\(x=0:2\)
\(x=0\)
b) \(\left(3x-1\right).\left(-\frac{1}{2x}+5\right)=0\)
=> \(\left(3x-1\right)=0\)hoặc \(\left(-\frac{1}{2x}+5\right)=0\)hoặc \(\left(3x-1\right)\)và\(\left(-\frac{1}{2x}+5\right)\)cùng bằng 0.
\(\orbr{\begin{cases}3x-1=0\\-\frac{1}{2x}+5=0\end{cases}}=>\orbr{\begin{cases}3x=1\\-\frac{1}{2x}=-5\end{cases}}=>\orbr{\begin{cases}x\in\varnothing\\2x=\frac{1}{5}\end{cases}}=>x=\frac{1}{5}:2=>x=\frac{1}{10}\)
a) \(5^x:\left(5^2\right)^2=625\)
\(5^x:625=625\)
\(5^x=5^8\)=> x = 8
Mấy câu kia tương tự
d) \(\left(x-1\right)^4-\left(x-1\right)^4\cdot\left(x-1\right)^3=0\)
\(\left(x-1\right)^4\cdot\left[1-\left(x-1\right)^3\right]=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^3=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x-1=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)
Vậy,..........
a) \(2^5.8^4=2^5.\left(2^3\right)^4=2^5.2^{12}=2^{5+12}=5^{17}\)
b) \(25^6.125^3=\left(5^2\right)^6.\left(5^3\right)^3=5^{12}.5^9=5^{12+9}=5^{21}\)
c) \(625^5:25^7=\left(5^4\right)^7:\left(5^2\right)^7=5^{28}:5^{14}=5^{28-14}=5^{14}\)
d) \(12^3.3^3=\left(12.3\right)^3=36^3\)
Dấu chấm là nhân nhé
minh anh sai rồi câu a là 2 mũ 17 mà ghi là 5 mũ 17
Bài 1:
a) \(\dfrac{10^6\times10^2}{1000^3}=\dfrac{10^{6+2}}{\left(10^3\right)^3}=\dfrac{10^8}{10^9}=\dfrac{1}{10}\)
b)\(\dfrac{625^2\times57\times3125}{25}=\dfrac{\left(5^4\right)^2\times5^5\times57}{5^2}=\dfrac{5^{13}\times57}{5^2}=5^{11}\times57\)
Bài 2 :
a) (x - 5)5 = (x - 5)10
⇒ x - 5 = 0 hoặc x - 5 = 1
⇒ x ∈ {5 ; 6}
a) 25 . 84 = 25 . (23)4 = 25 . 212 = 217
b) 256 . 1253 = (52)6 . (53)3 = 512 . 59 = 521
c) 6255 : 257 = (252)5 : 257 = 2510 : 257 = 253
d) 123 . 33 = (12 . 3)3 = 363
a)4^x=64
4^x=4^3
<=>x=3
b)3^x-1=27
3^x-1=3^3
<=>x-1=3
x=3+1
x=4
c)26+8x-6x=46
26+x(8-6)=46
26+2x=46
2x=46-26
2x=20
x=20:2
x=10
d)\(25\le5^x\le625\)
<=>\(5^2\le5^x\le5^4\)
<=>\(2\le x\le4\)
<=>\(x\in\left\{2;3;4\right\}\)
e)10+2x=4^5:4^3
10+2x=4^5-3
10+2x=4^2
10+2x=16
2x=16-10
2x=6
x=6:2
x=3
a) 4x = 64
\(\Rightarrow\)4x = 43
\(\Rightarrow\)x = 3
b) 3x - 1 = 27
\(\Rightarrow\)3x - 1 = 33
\(\Rightarrow\)x - 1 = 3
\(\Rightarrow\)x = 4
c) 26 + 8x - 6x = 46
\(\Rightarrow\)26 + 2x = 46
\(\Rightarrow\)2x = 46 - 26
\(\Rightarrow\)2x = 20
\(\Rightarrow\)x = 10
d) 25 \(\le\)5x \(\le\)625
\(\Rightarrow\)52\(\le\)5x\(\le\)54
\(\Rightarrow\)x \(\in\){ 2 ; 3 ; 4 }
e) 10 + 2x = 45 : 43
\(\Rightarrow\)10 + 2x = 42 = 16
\(\Rightarrow\)2x = 16 - 10
\(\Rightarrow\)2x = 6
\(\Rightarrow\)x = 3
a) 5.5x = 625
=> 5x+1 = 54
=> x + 1 = 4
=> x = 4 - 1
=> x = 3
Vậy x = 3
b) 8(x + 25) - 155 = 181
=> 8(x + 25) = 181 + 155
=> 8(x + 25) = 336
=> x + 25 = 336 : 8
=> x + 25 = 42
=> x = 42 - 25
=> x = 17
Vậy x = 17
c) (x - 9)4 = (x - 9)2
=> (x - 9)4 - (x - 9)2 = 0
=> (x - 9)2.[(x - 9)2 - 1] = 0
\(\Rightarrow\left[\begin{array}{nghiempt}\left(x-9\right)^2=0\\\left(x-9\right)^2-1=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x-9=0\\\left(x-9\right)^2=1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x-9=0\\x-9=1\\x-9=-1\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=9\\x=10\\x=8\end{array}\right.\)
Vậy \(\left[\begin{array}{nghiempt}x=9\\x=10\\x=8\end{array}\right.\)
a)5*5x=625
\(\Rightarrow5^{x+1}=625\)
\(\Rightarrow5^{x+1}=5^4\)
\(\Rightarrow x+1=4\)
\(\Rightarrow x=3\)
b)8(x+25)-155 =181
\(\Rightarrow8\left(x+25\right)=336\)
\(\Rightarrow x+25=42\)
\(\Rightarrow x=17\)
c) ( x-9)4 = ( x-9)2
\(\Rightarrow\left(x-9\right)^4-\left(x-9\right)^2=0\)
\(\Rightarrow\left(x-9\right)^2\left[\left(x-9\right)^2-1\right]=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}\left(x-9\right)^2=0\\\left(x-9\right)^2-1=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-9=0\\x-9=\pm1\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=9\\x=10\\x=8\end{array}\right.\)